$Q = mc\Delta T$

where:

• $Q$ = heat transferred (J)
• $m$ = mass (kg)
• $c$ = specific heat (J/kg·°C or J/kg·K)
• $\Delta T$ = temperature change (°C or K)

Common Specific Heats:

Water has the highest specific heat of common substances! This is why:

• Coastal areas have moderate climates
• Water is used in cooling systems
• Land heats/cools faster than oceans

Sign Convention

• $Q > 0$: Heat absorbed (temperature increases)
• $Q < 0$: Heat released (temperature decreases)

$\Delta T = T_{final} - T_{initial}$

If $\Delta T > 0$: temperature increases → heat absorbed If $\Delta T < 0$: temperature decreases → heat released

Calorimetry

Calorimetry measures heat transfer using conservation of energy.

Principle: In an isolated system, heat lost = heat gained

$Q_{lost} + Q_{gained} = 0$

or

$\sum Q_i = 0$

Typical Setup:

• Hot object placed in cold water
• System isolated (insulated calorimeter)
• Final temperature measured
• Energy conservation: $Q_{hot} + Q_{cold} = 0$

Heat Capacity

Heat capacity (C) is heat needed to raise an object's temperature by 1°C:

$C = mc$

$Q = C\Delta T$

Units: J/°C

Difference:

• Specific heat (c): Property of material (per kg)
• Heat capacity (C): Property of specific object (total)

Methods of Heat Transfer

1. Conduction

• Heat transfer through direct contact
• Molecular collisions
• Requires material medium
• Example: Metal spoon in hot soup

2. Convection

• Heat transfer by fluid motion
• Hot fluid rises, cold sinks
• Requires fluid (liquid or gas)
• Example: Boiling water, ocean currents

3. Radiation

• Heat transfer by electromagnetic waves
• No medium required (works in vacuum)
• All objects emit thermal radiation
• Example: Sun warming Earth, heat lamps

Problem-Solving Strategy

1. Identify all objects exchanging heat
2. Set up energy conservation: $\sum Q = 0$
3. Write Q for each object: $Q = mc\Delta T$
4. Determine signs:
   • Heating: $Q > 0$, $\Delta T > 0$
   • Cooling: $Q < 0$, $\Delta T < 0$
5. Solve for unknown (usually final temperature or specific heat)
6. Check reasonableness: Final T should be between initial temperatures

Common Mistakes

❌ Using wrong sign for heat (lost vs. gained) ❌ Forgetting to convert units (g → kg, cal → J) ❌ Using Celsius in place of Kelvin inappropriately (ΔT is same, but not absolute T) ❌ Neglecting heat lost to surroundings (real calorimeters aren't perfectly insulated) ❌ Mixing up specific heat (c) and heat capacity (C)

Find: Heat required $Q$

Solution:

Step 1: Calculate temperature change. $\Delta T = T_f - T_i = 80 - 20 = 60°\text{C}$

Step 2: Apply heat formula. $Q = mc\Delta T$ $Q = (2.0)(4186)(60)$ $Q = 502,320 \text{ J} = 502 \text{ kJ}$

Answer: 502 kJ of heat is required

This is equivalent to about 120 food Calories (kcal).